Irodov 1.109: When A Circular Orbit Resists Nudging – Amit & Deepali Sinha Foundation

Problem Irodov 1.109

A particle moves in a plane subject to a force whose direction is always perpendicular to the particle’s instantaneous velocity, and whose magnitude depends only on the particle’s distance $r$ from a fixed point $O$ in the plane, as $f(r)=k/r^n$ for some constants $k$ and $n$ . The particle can move on a circle of some radius $r_0$ centered at $O$ , with the force supplying exactly the centripetal pull needed. Find the values of $n$ for which this circular motion is steady.

A particle moving on a circle of radius r_0 about the fixed point O. The force F(r) is perpendicular to the velocity v, which is tangent to the circle, so here F lies along the radius. — A particle moving on a circle of radius $r_0$ about the fixed point $O$ . The force $\vect F(r)$ is perpendicular to the velocity $\vect v$ , which is tangent to the circle, so here $\vect F$ lies along the radius.

Given

$f(r)=k/r^n$ , force always $\perp\vect v$

Find

the values of

n

for which the circular orbit is stable

Solution

Step 1Equations of motion in polar coordinates

We read “steady” as meaning the circular orbit is stable against a small radial perturbation: a small radial disturbance stays small rather than growing. With $O$ as pole, write $v_r=\dot r$ , $v_\theta=r\dot\theta$ , and $v=\sqrt{v_r^2+v_\theta^2}$ for the speed. A force of magnitude $f(r)$ perpendicular to $\vect v$ , rotated $90^\circ$ from it, has components

F_r = -f(r)\,\frac{v_\theta}{v}, \qquad F_\theta = f(r)\,\frac{v_r}{v}.

Since $\vect F\cdot\vect v=0$ identically, $d(v^2)/dt=0$ , so the speed $v$ is constant along any trajectory the particle actually follows; call its value $v_0$ on the orbit under study. Newton’s second law in polar form then reads

\begin{aligned} m(\ddot r - r\dot\theta^2) &= -f(r)\,\dfrac{v_\theta}{v_0} \\ m(r\ddot\theta+2\dot r\dot\theta) &= f(r)\,\dfrac{v_r}{v_0} \end{aligned} \tag{1}

Step 2The circular orbit itself

On the circle $r=r_0$ , $v_r=0$ and $v_\theta=v_0=r_0\dot\theta_0$ (constant), so the radial equation reduces to a balance between the centripetal term and the force.

-m r_0\dot\theta_0^2 = -f(r_0)

f(r_0) = \dfrac{mv_0^2}{r_0}

(2)

Step 3Linearizing a small radial perturbation

The construction this step linearizes about: the particle is nudged off the circle of radius r_0 onto the neighboring radius r_0+x (dashed), with x small. — The construction this step linearizes about: the particle is nudged off the circle of radius $r_0$ onto the neighboring radius $r_0+x$ (dashed), with $x$ small.

Let $r=r_0+x$ with $x$ small, so $v_r=\dot x$ is first order and $v_\theta=v_0+O(x^2)$ . The angular rate then follows from $v_\theta=r\dot\theta$ , with the reciprocal radius expanded to first order in $x$ .

\begin{align*}\dot\theta &= \frac{v_\theta}{r} \approx \frac{v_0}{r_0+x} \\ &= \frac{v_0}{r_0}\left(1+\frac{x}{r_0}\right)^{-1} \\ &\approx \frac{v_0}{r_0}\left(1-\frac{x}{r_0}\right)\end{align*}

The centripetal term of the radial equation follows by squaring this and multiplying by $r=r_0+x$ , dropping every product of two small quantities as it appears.

\begin{align*}r\dot\theta^2 &\approx (r_0+x)\,\frac{v_0^2}{r_0^2}\left(1-\frac{x}{r_0}\right)^2 \\ &\approx \frac{v_0^2}{r_0}\left(1+\frac{x}{r_0}\right)\left(1-\frac{2x}{r_0}\right) \\ &\approx \frac{v_0^2}{r_0}\left(1-\frac{x}{r_0}\right) \\ &= \frac{v_0^2}{r_0}-\frac{v_0^2}{r_0^2}x\end{align*}

Expanding $f(r)\approx f(r_0)+f'(r_0)x$ and substituting both expansions into the radial equation of (1), using $v_\theta/v_0=1+O(x^2)\approx1$ ,

\begin{align*}m\ddot x - m\left(\frac{v_0^2}{r_0}-\frac{v_0^2}{r_0^2}x\right) &= -f(r_0)-f'(r_0)x\end{align*}

The zeroth-order terms cancel by (2), leaving the linear equation for the perturbation

m\ddot x + \left[\dfrac{mv_0^2}{r_0^2}+f'(r_0)\right]x = 0

(3)

Step 4The stability condition

Differentiating the force law and substituting the balance condition (2) for $f(r_0)$ ,

\begin{align*}f'(r_0) &= -n\,\frac{f(r_0)}{r_0} \\ &= -n\,\frac{mv_0^2}{r_0^2}.\end{align*}

Substituting into the bracket of (3),

\begin{align*}\frac{mv_0^2}{r_0^2}+f'(r_0) &= \frac{mv_0^2}{r_0^2}-n\frac{mv_0^2}{r_0^2} \\ &= \frac{mv_0^2}{r_0^2}(1-n)\end{align*}

This is positive, giving stable oscillation about $r_0$ , exactly when $1-n>0$ .

n < 1

(4)

Check

At $n=0$ the force is uniform in magnitude ( $f=k$ , independent of $r$ ), and $1-n=1>0$ confirms stability, matching the intuition that a constant-magnitude, always-perpendicular force is the textbook case of steady circular motion. At $n=1$ the bracket vanishes exactly, the marginal case where a radial nudge neither grows nor oscillates back, consistent with $n<1$ being a strict inequality. Negative $n$ is admissible too: it just means $f(r)=k/r^n=k\,r^{|n|}$ grows with $r$ rather than falling off, and $1-n>1>0$ then holds automatically, so those orbits are stable as well, the least obvious part of the corpus’s own answer.

Remark

This $n<1$ threshold is narrower than, not more permissive than, the ordinary central-force stability condition $n<3$ for a force $f(r)=k/r^n$ directed straight along $\vect r$ (an inverse-square or weaker attractive law): every $n$ that keeps this orbit stable also keeps a true central orbit stable, but not conversely (take $n=2$ , stable centrally, unstable here). The difference is the extra tangential force component $F_\theta=f(r)v_r/v_0$ present here and absent for a true central force: it exists only while the particle is perturbed off the circle, and it destroys the angular-momentum conservation a true central force’s stability analysis normally leans on, which is why the window narrows.